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Q.

A 20 mL sample of vinegar of density 1.06 g/cc requires 43 mL of 0.40 N $N a O H$ for complete neutralization. The percentage by weight of acetic acid, $C H_{3} C O O H$ , in the vinegar is_____.

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answer is 4.83.

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Detailed Solution

$20 \times M_{C H_{3} C O O H} = 43 \times 0.4$

$\Rightarrow 20 \times \frac{10 \times x \times d}{60} = 43 \times 0.4$

$\Rightarrow 20 \times \frac{10 \times x \times 1.06}{60} = 43 \times 0.4 \Rightarrow x = 4.867$

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